PHY303k_-_Spring_2010_-_HW_10 - saldana (avs387) Homework...

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Unformatted text preview: saldana (avs387) Homework 10 florin (58140) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A weight lifter lifts a mass m at constant speed to a height h in time t . What is the average power output of the weight lifter? 1. P = mg h 2. P = mg ht 3. P = mh 4. P = g 5. P = mg h t correct Explanation: The amount of work to lift a mass m to a height h is mg h . The power is the amount of work done divided by the time t . Alternately, power has units of energy over time. The only answer with the correct units is P = mg h t . 002 10.0 points Units of power include which of the following? I) Watt II) Joule per second III) Kilowatt-hour 1. I and II only correct 2. III only 3. II and III only 4. I, II and III 5. I only Explanation: [Power] = [Energy] [time] Watt = Joule second is the unit of power. Kilowatt-hour is a unit of energy. Thus, only I and II are correct. 003 10.0 points A car weighing 7400 N moves along a level highway with a speed of 75 km / h. The power of the engine at this speed is 62 kW. The car encounters a hill inclined at an angle of 6 . 9 with respect to the horizontal. If there is no change in the power of the engine, and no change in the resistive forces acting on the car, what is the new speed of the car on the hill? Correct answer: 57 . 7488 km / h. Explanation: Since the velocity is constant, the car must be in equilibrium. Assume F is the pulling force provided by the engine and f is the fric- tional force. On the level highway F 1 = f F 1 P v On the hill: F 2 = f + mg sin( ) v 2 P F 2 = F 1 v F 2 = F 1 v F 1 + mg sin...
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This note was uploaded on 08/30/2010 for the course PHY 58140 taught by Professor Florin during the Spring '10 term at University of Texas at Austin.

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PHY303k_-_Spring_2010_-_HW_10 - saldana (avs387) Homework...

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